Question

Por favor é urgente
Se a = 2 arc cos (-1/2), então tg a é igual:

Answer 1

$\large\boxed{\begin{array}{l}\sf a=2~arc~cos\bigg(-\dfrac{1}{2}\bigg)\\\sf arc~cos\bigg(-\dfrac{1}{2}\bigg)=x\implies -\dfrac{1}{2}=cos(x)\\\sf a=2x\\\sf cos^2(x)=\dfrac{1}{4}\\\\\sf sen^2(x)=\dfrac{4}{4}-\dfrac{1}{4}=\dfrac{3}{4}\\\sf sen(x)=-\sqrt{\dfrac{3}{4}}=-\dfrac{\sqrt{3}}{2}\\\sf tg(x)=\dfrac{sen(x)}{cos(x)}\\\sf tg(x)=\dfrac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}}=\sqrt{3}\\\sf tg(a)=tg(2x)=\dfrac{2tg(x)}{1-tg^2(x)}\\\sf tg(a)=\dfrac{2\sqrt{3}}{1-(\sqrt{3})^2}\end{array}}$

$\large\boxed{\begin{array}{l}\sf tg(a)=\dfrac{2\sqrt{3}}{1-3}\\\\\sf tg(a)=\dfrac{\backslash\!\!\!2\sqrt{3}}{-\backslash\!\!\!2}\\\sf tg(a)=-\sqrt{3}\end{array}}$